91Ó°ÊÓ

Given equation is$y^{\text{'}\text{'}}−6y^{\text{'}}+9y=12x{e}^{3x}$

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one uses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

The given differential equation is

$\begin{array}{l}{y}^{\text{'}\text{'}}−6y^{\text{'}}+9y=12x{e}^{3x}\\ (D^2−6D+9)y=12x{e}^{3x}\end{array}$

The auxiliary equation can be written as

$\begin{array}{c}{m}^2−6m+9=0\\ m^2−3m−3m+9=0\\ m(m−3)−3(m−3)=0\end{array}$

The roots of the equation are

$⇒m=3,3$

$C.F=(C_{1}x+C_2)e^{3x}$

Now find the P.I.

$\begin{array}{l}y=x^2{e}^x(Ax+B)\\ y^{\text{'}}=x{e}^{3x}\left(3Ax\right(1+x)+B(2+3x)\\ 3Ax+B=6x\\ A=2\text{andB}=0\end{array}$

The obtained results are

$\begin{array}{l}\text{P.}I=2x^3{e}^{3x}\\ \text{C.}S=(C_{1}x+C_2)e^{3x}+2x^3{e}^{3x}\end{array}$

The solution of the differential equation is.$y\left(x\right)=(C_{1}x+C_2)e^{3x}+2x^3{e}^{3x}$